UVA1626-Brackets sequence（动态规划基础）

Problem UVA1626-Brackets sequence

Time Limit: 4500 mSec

Problem Description

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs. The input file contains at most 100 brackets (characters ‘(’, ‘)’, ‘[’ and ‘]’) that are situated on a single line without any other characters among them.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line. Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.

Sample Input

([(]

Sample Output

()[()]

题解：这个题挺好的，区间dp，可以写成记忆化搜索，容易忽略的地方是如果区间两边的括号匹配，那么可以用中间的部分转移，然后就是普通的分成左区间和右区间进行转移，这个题比较有价值的地方在于打印解的过程，应该学习一下，就是根据结果，逆推回去，这个方便在不用中间记录转移路径，代价就是时间上会有额外的开销，不过一般不至于因此就TLE，因为解一般很少。输入输出有坑，需要用fgets，并且注意fgets会把最后的' '读进来，因此真实串的长度需要-1.

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int maxn = 100 + 10;
 6 
 7 int iCase, dp[maxn][maxn];
 8 char bra[maxn];
 9 bool vis[maxn][maxn];
10 
11 bool match(char a, char b) {
12     if ((a == '(' && b == ')') || (a == '[' && b == ']')) return true;
13     return false;
14 }
15 
16 int DP(int l, int r) {
17     if (dp[l][r] >= 0) return dp[l][r];
18     if (l == r) return dp[l][r] = 1;
19     if (l > r) return  dp[l][r] = 0;
20 
21     int &ans = dp[l][r];
22     ans = r - l + 1;
23     if (match(bra[l], bra[r])) {
24         ans = min(ans, DP(l + 1, r - 1));
25     }
26     for (int k = l; k < r; k++) {
27         ans = min(ans, DP(l, k) + DP(k + 1, r));
28     }
29     return ans;
30 }
31 
32 void ans_print(int l, int r) {
33     if (l > r) return;
34     if (l == r) {
35         if (bra[l] == '(' || bra[l] == ')') {
36             printf("()");
37         }
38         else {
39             printf("[]");
40         }
41         return;
42     }
43 
44     int &ans = dp[l][r];
45     if (match(bra[l], bra[r]) && ans == dp[l + 1][r - 1]) {
46         printf("%c", bra[l]);
47         ans_print(l + 1, r - 1);
48         printf("%c", bra[r]);
49         return;
50     }
51     else {
52         for (int k = l; k < r; k++) {
53             if (ans == dp[l][k] + dp[k + 1][r]) {
54                 ans_print(l, k);
55                 ans_print(k + 1, r);
56                 return;
57             }
58         }
59     }
60 }
61 
62 int main()
63 {
64     //freopen("input.txt", "r", stdin);
65     scanf("%d
", &iCase);
66     while (iCase--) {
67         fgets(bra, maxn, stdin);
68         memset(dp, -1, sizeof(dp));
69         int len = strlen(bra);
70         int ans = DP(0, len - 2);
71         ans_print(0, len - 2);
72         printf("
");
73         if (iCase) printf("
");
74         fgets(bra, maxn, stdin);
75     }
76     return 0;
77 }